Question

What is the maximum value of the force F such that the block shown in figure does not move

• A. $20N$
• B. $10N$
• C. $12N$
• D. $15N$

Answer 1

Answer: (A)

$20N$

For vertical equilibrium of the block

$N=mg+Fsin{60}^0$

$=\sqrt{3g}+\sqrt{3}\frac{F}{2}$

For no motion, force of friction

$\begin{array}{c}f\ge F\cos {60}^{0}or\mu N\ge F\cos {60}^0\\ \frac{1}{2\sqrt{3}}\left(\sqrt{3g}+\sqrt{3}\frac{F}{2}\right)\ge \frac{F}{2}\\ \Rightarrow g\ge \frac{F}{2}\\ \Rightarrow F\le 2gor20N\end{array}$

Therefore , maximum value of $F$ is $20N$.